chap003

# chap003 - .EICCtronic Circuit Analysis and Design...

This preview shows pages 1–19. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: .EICCtronic Circuit Analysis and Design. 2’"1 edition Solutions Manual Chapter 3 Exercise Solutions 53.1 ﬂglja 0900 Fora=0.980.ﬁ=1—_'bm=49 Fara=0.995.ﬁ=1;-:':%=i99 49<5<199 E32 a=ﬁ€3=g=m HES-=st 16 E33 7 IE=U+J3JIR So(l+ﬁl=f,—:=E%=SL2S=> ﬁ=80.3 a=liﬁ=g%=>a=0.98?7 Ic = Na = [30.3)(9.00 0A) => rc = 0.7T1mA E14 a 0.990 5- hymns-99 I I5 -— “5° => [5 .—. 21.50 M 9 = (1+5) " 100 I: = arr; = {0.590)(2.i50) =:~ lg = 2.13 mA 53.5 11-329 rn‘Ic— Ic Ic=0.l mha-ra =1.5 MR 15: 1.0 mAaro=150 kn Ic=10mA=1rra=15kQ 1 Ic—lon l+ﬁ ) => Io = 0.9363 mA A! ch =10, [c = [0.9868)(l + a I —- 1.12 link i 10 0| b ‘4 150 1': =1 =;u(1+ _—_-, Io =0.9934 TBA A! Vcs :10. {C =(0.9934)(1+ ___, 150 1’: =1.06 m.-'\ 53.7 . EV a 200 _ Bbczo = —:‘/%—u = W = 40.0 volts E33 BV BVCEO = C311 :13 va =(U100)(30)=139v 53.9 a. V.=C|.'2<V33[Oll}=’fﬁ=!c=0‘VO=SV P = D b. V.- = 3.6 Transismr is driven into saturation. 3.5—0‘ - Ia— ﬂ-vﬁqlal =4 3111A 5- ch[3at} 5—0.2 _ _ :9. I =10.9 A Ir: RC OM 5—11}... Now that I—C = “,3 = 2.41 < 5 which shows 15 4.33 [hat «in: transistor is Ludccd in mmran'nn. P = [cvcg 4- Java; = (10.9)“).2) + (4.53MOJ) = 2.18 + 3.17 P = 5.35 In“: 153.10 For Vac = 0 =: Va = 0.? V Thcnlc=5gl57=\$1c=9JTmA _[c_9.TT_ _ andia— a _ so _o.19amA v; = I503 + vuum) = (0.195}(0.54) + 0.7 = V: = 0.525 V Powc-r = Ich-z + IEVBE = [9.77)(0.7) + L0.195)(0.7) Em“ = ﬁ mW Electronic Circuit Analysis and Design, 2"‘1 edition solutions Manual E3.“ For v; =4 V and {Ca =1.5 mA RC2 => Rr=4kn I _ -V35(°n] -[—10¥ E '- RE 101‘ = —-- = .'1' . 15 (‘00 IC 1.) a m-\ RE:£‘—.°—;'—°-=Rg=s.mm 1.313 —“— E3.l2 lD-Vc _10 —s.34 _ Ic- — RC — 4 => 1c,- -— 0.910 m.—‘\ I _ —V55(0n)-[-1m_l0 -0.7 2, E " R5 ‘ 10 1'} =0.930 [:13 1c 0.915 Ic—alg=>ar—.-I—E-—olgm =3-Cr—D. 8.1-9 I5 = I; —[5- =0.930 — 0.915 5' = mA I: 0.915 ‘3’ [a ‘0015 93““ Va; = Vc -- V; = 5.3-1- (—0.70) => VEE =7.04 V 53.13 _ 10—V55[on) __ 10 —D.T Is- =>IE=L16mA RE 8 15 1.15 = = —-—-— = . 'r' 15 (1+3) 51 #13 22 11A. :3 so _ [c — 1 +513 -— ﬁ(1.15)=ﬂ- IE — Lli [TIA Vc = IcRc — 10 = (1.14)(4} -- 10 = -5.44 Vgc = 0.7 -— (-5.44) =9- Vﬁg = 5.14 V 53.14 IE _ V35 - Vsahn) _ RE ‘I -G.T =>R5= 1° :R5=3.3kﬂ I: = of; = (0.9920)(1.DJ =5 1c = 0.992 mA I; = I; — IC = 1.0 — 09920 = Ia = 0.0030 [119. Vac = -Vsc -—- 1cm: — Vcc = (0.9921(1} -— 5 =’ Vic = 4.01 V E3J5 v.93 = IBRB + Vagton} + [53.5 a fgﬂy + V3,;(onj4- (1 +H'II5R5 I, = ﬂ = i Ra +(1+ 31R; 10 + [76)(1) =3+ In = 15.1.HA Ic- = 315 =(TS}(15.1 HA) ==> [g = 1.13 mA I: =l1+ SUB =(T5}[15.1u.»\)=:» M Vcs = Vcc + V93 - IcRr: - IERE = 8+ '3 - (1.13)(2.5'| —— [1.153(1) Veg = 6.03 V E116 VICE =15 =v V5 = 2.5 V = 15R; V23 = 1335 + Vaswnl + V5 Ia _ V33 - Vadon) — V5 __ 5 — 0.1 — 2.5 R3 "' 10 I3 = 0.15 mA => I; = (101)(0.18) => I; =15.18 mA 2.5 S R = = , = o 5 ls‘laﬁﬂg 0138 m 1389 E317 V55 = 1535 + V£B(0n} + [3R3 2. 13:12 mA=~I5~= £20.0431mA _ I3 .. ﬂ _ Ic —— (1+ﬁ)ls— (51)(2.2) :- Ic -2.16 mA v3... = [2.21m + 0.7 + (0.0431)(50) ==> V53 = 5.05 V Vgc = 5 - IERE = 5 - {2.2](l) =- Vgc = 2.3 V Ella [1) E = IERB + Vas(on]+ I53; (2} 5 = IcRc +Vc5(sa.1) +IER5 11-: = 15 + Tc (1) 5 =1013 + 0.7 + (15 + 1cm) (2) 5 = 4!: + 0.2 + (Ia + Ic)[1) (1} [5.3 = I: +11I3]x s —- 26.5 = SIC + 5513 (3)4.8=51c+I5 4.8=5Ic+ [a 21.7 = 541-3 :5 [g = 0.402 mA From [1). IC 2: 5.3 — 1113 a I; .—. 0.380 mm £3.19 V55- : Vgc-{Sln = 0.2 V H —B.2 - ("5] _ 5 — 0.2 Ic "T 10 =? I; =U.43 mA 0.48 Ia=%‘=—, —w_=_g : v, = —(0.24}{20) — 0.1 => vE = —s.5 v lest Your Understanding Chapter 3: Exercise Solutions E320 E322 I, V; = —4.5 V = V55 < Magma) => Transis‘mr is 3' L3 = b} =_ U‘ h" :13: = {m =1“ = I” = D cutoﬁ'. IE=IQ=IE=CL V£E=10V ‘1' =3? . b. V: = —3.5 V Transismr is active. '3‘ h = a x" Ll! = 0' IE” = In 2 0 5 —0.T _ V; = IBRB + V35[on) + IERE — 5 [m _ 0‘95 :13] .- 4”; mA ‘ _ '3 5 — 3.5 = IBHD} + 0.7 + (75}1301) la: a 0 and. 21c] 2h; = 3 mA 5-15—03: . '. = .__..___.._._ = I - 1» =02 \ 1; “WWW! 00025.: mA 0 #IE=Z_53#A c, V1=Vg=5‘~'-. IE§=JI32=4.5311L‘\ I: = ﬁIg = {75)(255 3A} => 1.: 7: 0.191mA In = 8 mA. 161 =16: = 4 IRA! VB = 0-3 V I; ={1+ 5)]; = (76)I:1.55 MA) E3 23 = IE = 0.194 mA I . VCE=10-IcRc-I5RE v0-5—rcRc=S—ﬁsRc and =10 —(o.191}{2}-(o.194)(4) i = V5,, +Av, —V,E(on) Va; = 5.34 v " R3 Then c. V. = +3.5 V Transistor is in saturation. Av _ _ﬂRcAvl a _..__........ R (1} 3.5:13R3+V55(on)+IER5—5 or " [2) 5 = IcRc + VCEBM} + J'sz — 5 Ave _~ (3) IE =13+Ic Av, Ra Let ﬁ=100, Rc=5kﬂ, Rﬂ=100kﬂ (1‘13.5+5—D.7=1DI&+4(I5+1c) Then [2) 5 + s — 0.2 = 2n: + ms + :c) Ava ‘ -(100)(s) _ S (1) 7.3 = 1415 4.41,,- Av, ' 100 ' [2} 9.3 = 415 + EIc Want Q-paintto be 3 x [1) =- 214 = 4213 + 121C "0(Q-Pl)=25=5_(100)‘{39(5) Then me =- 19.5: SIE+12J§ V -07 =0.005mA I =0005=L 3.3 = 3“? .LL—‘ “9 100 =I§ =D.li'2 mA (I); V =12 7.8 =14(U.112}+ ﬁfe = 1.553 + 416 "L a I = L56 mill Also 1Cg =13!” =(1oo)(o.oos) [c "_ 0r E:l\$.9<6=¢1nsaruratiun. ICQ=05mA IE=I3+Ic=s-f§=1.57mA E32 Vcs = 1"c-,ns:(5a.r.)= 0.2 - 4 a. For Vcsq = 2.5 V 5 [ca = 5 -22-5 =:- IcQ =1.25 mA 5-15-02 I sat ==———-= = 9 c( ) R 15::- R 0.22014) hm = I? = ___1l-D-05 z, 15;, = 12.5 pA 1c 15 5—0.3 ' - _ ~ Itz—02—0=0.7SMA= R, Thgnﬂszznlozgﬁﬂﬁ=344kn or . RB :55“) b. 159 15 indcpcndcm of 13. For¥c5q=1V, Ic=3_1=2mA Electronic Qircuit Analxgis and Design, 2"d edition Solutions Manual 1c 2 - = __ = 1 =150 _ 1-0-r _ '3 In .0125 a 3 130 —- = 139 —17.6 #3 . . 5 - 4 _ . For ‘11ch = 4 V, 1c = ., = 0-0 mA Icq = 131'“ = (35)(17.6 11A] L: 0.5 :3 1 =1.32 11114 := —— = 3 = 0 3.9—— ‘5 13 0.0125 =" “ 120 = (1 +1311“ = (?6](l7.6 11A) => IEO =1.34 111A E325 Vase = 5— (1.321(1) — (1.3410121 5040<d<160 ._0-- =>V55Q23AIV ° 800' = In = 0.005355 mA ———*"—" 13 = 75 =5 {an = (751100053751: 0.403 111.1 E317 :3 =150 =1- Icq =1150)(0.0053751= 0.606 11131 Largest Icq = 5111311051 Vcsq 5 —- 1 = \$.96 = 0.506 m 9-: ‘K 4 ° ' = 2.45 m 0.405 I39: ‘21:!“ 3:150:35: S=TS=~RC= For a nominal Ic = 111.604 mA and 17.35.; = 2.5 5 — 2.5 RC ‘ 0.604 For Ice = 0.403. Rfagx chq = 5 — (0.403)(4.11) = 3.33 V For Icq = 0.606. Vega = 5 - (0.306)[1.14] = 1.66 v 50 fox R: = 4.14. 1.66 V 5 Vcaq 5 3.33 V = 4.14 = R: 1' R; + R: =11.25 kn. 5 =15!) Ic z Ia. Vega = 2.5.V So Icq :..- 15.; = = 2.081mA E326 [so = 16?" = ﬁg =13.9 1.111. n. Rm = Rlﬂﬁz = 9||2.25 =5 Rm =16 1:11 I _ V” _VE£[W) V R: “5 5" Rn: +11 + 51R: 3'” = (m) V” = [9-1- 2.25)“) : -u'——V = 1‘0 v (*RITEZ) 'cc - V5503“) 0.0139: 1 2 h. I“ = v” — Vezfan) z 1— 11.7 m +11 +5135 Rm +11 + 15mg 1.6 + (151110.21 R; )6) * M =# Isa = 9.375 111A _—_ Riggs Ice = 31's.; = (150149.375 11A) ([125] +(1511(0»2) =>Ico = 1.11 mA I30 ={1 +13)I\$Q 515:: = 1.42111}. Vcsq =5-Ichc-IEQRE R2 = 11.25 —- R1,. 50 0.0139[R;(11.25 - R;]+ (15111021111251) = 5 —(1.»111(1) 41.421012) V = 3.31 V , .. -——'9————” =a(11.25—R1) 40.111125) .3 pm- ,3 = 75 0.15631 — 0.0119113 + 1.72 = 56.25 — 511, — 7.575 0.01393? —- 5.15611. + 43.66 = 0 =53; - (0.1)(11251 5.156 i (5.156)’ — 4(0.01391{43.561 _ 2(00139) => 34: 8.67 kn wd R1 = 2.55 m R-l Test Your Understanding Chapter 3: Exercise Solutions E23 .28 I dc equivalent circuj: :3 =150. Rn: = RIHRZ. Vary = (—13—)(1m—5 RI+Rz 5—0 1:4: 1 =5m.'-\ fax; 5 I =—=—-—-_- ‘u so a 150 0333mA VTH-t-S-UJ 159- ‘ Err-1+0 +5)Rz Sc: RrH = (0.1|(1 +.5)Rs I __ V111; +4.3 3° ' (1.1)(1 +51Rs R: _ (R: + 32):”) — 3+Jt3 : [10333 = (1.1)(151n02) (0.0333)(1.1)(151)(0.21=(%D(1n)— 0.7 R; _....._—. = 0_ 30 (R1 + R2) 1806 Rm — T— : {0.1p{151}(0.2} = 3.02 m Then R.{0.1006) = 3.02 =- 121 = 15.7 m a, = {0.1805}(16.T + R1} = 0.31903; = 3.02 :9 R; = 3.65 M! 53.29 :3 =120. Vcsq = 5- V . R: RTE R1llﬁe. TH (121+ R2){ID] 5 In; == Isa 10 - Vcsq 3-: + R: “3-5 1:4 — =>- Icq = 3.33 mA 50 Ice = Isa: LEE-=— %:—:-=? {30 =0-0273 "‘A v” +5 -0.7 .r =——-——-— Ba Rn: +(1 +5135 .56! RTH = (0.1)“. +3313; R: . — —D.-' (R1+R1){10) 5+5 [1.13(121)(0.3) 1'90 :— D.0278 = [0.027a)(1.11(121)(0.3} =( R” )(101— 0.7 Ri+R2 (-—£’—)—0131 RI+RI "- . RIR: _ _— _- . = 0.! l . R73 R1 R2 leﬂ 131) ( 21)“) 3] =:- R; = '20.} in R: =(0.131)(20-1+ Ra] =:» 0.01%; = 3.03 m ﬁ' R2 = 4.44: kg £3.30 5 “ {100 = = — = .99 A a- fag (1+E)IEQ \10])(1} U m 1 mA 1 = —--— = — .90 . [5" 1+3 101:"9 “A '5 = --I3qR3 = —[0.UU§9)(50} ==> Va = —D.495 V . _ . Icq _ I 0.99 x10"' 03,; _ vr1n( IE ) _ [0.0263111 («———-s x 10m ) =. V55 = 0.530 V v5 = V3 _ V55 = 43.495 _ 0.530 =9- V; = “1.13 V v.5 = 10 - (0.993(3) = 5.05 \' Vcaq = W: — V; = 5.05 — (-1.13; 201-5}; = 0115 v b. [sq = 1111A. 1-3 = 51—] =U.0195 mA v5 = 400195150] = —0.93 v = V5 13.; = (553%)(1) = 0.95 mA 0.90 x 10-3 3 x [0"“ V; = -D.95 - 0.629 = -1.61 vs =10 - (0.95)(5) = 5.1V Vcsq = 5.1 — (—1.61) =- chq = 5.71 v V55 = (0.026|1n( ) = 0.629 \' Electronic Circuit Ana} sis and Desi 2"“ edition Solutions anual E331 E133 ._ .33.. _ IS. .. ’_° _ {a _1+ _ 121,113 .. (121)120) _ 11310155} 1/E = IQ10.105) +0 1 20 [co =(1fﬂ)rsq = In -—- (0.992110 VC 2 ICQRC — 5 = {0.992)IQ(4) — 5 = 3.97Iq - 5 VECQ 2 VS — VC = [19(0105) + 0.7] — {3.9719 — 5] =‘ 4.80512; + 5.7 — 3.80511; + 5.7 = 3 . =0 IQ = 0.710 mA I _ VrH - VBE(°“} _ 4 ___u'7 E" ‘ 12m +(1+31R21 ‘ 33.3 +(101'1121 131:14 0101. 1C; =1.40 1119.. M. E332 v3: = 4 —(0.011)(33.3) .51: 45V :9 E] = 3.53 V. V5; = 2.83 \" IR+IB? =Ic1 12 —- Va 12 -(Vc1+0.7] =’ 5 + (1011(2) =1'“ 12 (12—07) _h Va. ‘5"+ 11011121 “"w‘ 5 +(101)(2) 2.4 + 0.0559 — 1.40 = meq + 0.00495) VEI = Va; = 5.15 V. VE; = 5.35 ,, _ - [5: = :5 [52 = 3.105 mA. Ic; = 3.04 mA. 132 = 30.4 11.4 Va; = (3.041115) :3 11-, = 4.55 V R7}; = SOHIOCI = 33.3 kn . 50' - VTH —- (m)(10) —- a — —-1.67 V - - 1.67 - 0.7 9.63 I = -a——-—---—- = :— = .3 . 9' 33.3+11011(21 235 1‘ “" Icl = 1.12 mA. 1'51 = 1.13 1112‘. V31 = IE]. H5: — 5 ={l.13}(2} — 5 = —2_74 V ch = 3.25 V =- Va; = 0.51 V =- Vgg = 0.51+O.T =1.21 V - — .21 IE, :7 =1.90 m}; =3 1‘3, = 13.3 11A 1.3, = 1.30 111.”. Im = In — 1m =1.12 - 0.01311 = 1.10 mA 5 —0.51 RC1 —- =9 RE; —l.DB kn VEc: = 2.5 =- Vc: = V17: - VEC: = 1.21 - 2.5 = —1.29 -1.29- (—5) Re: = L83 =5- Rg: 21.9? 111'! Electronic Circuit Analysis and Designl 2"“ edition .___________501Uti0“5 Manual Chapter 3 Problem Solutions 3.l 3.2 (a) Patti—2110: G=L=m=099099 1+.5 111 ' Forﬁ=180: 180 G — m — 0.99915 0.99099 < a < 0.99448 (b)!C =ﬁ1,=ll0(50p4):> 1c =550mA or I: =180(50p,4)=>. Ic = 9.00m SD 5.50 s [c s 9.0 mA 3.3 5=ICR+VCE - V35=Ic(2]+1-0.7 =:~ Ic = 2.35 mA Ic 2.35 15— a —m=>fg=2.39m2°u 3.4 Same Figure as Problem 3.3 5—1.3 vc =—0.7+1=1.3V', ic=—1-——: it. =l‘85mA ' LBS . For ﬁF=120, :B=—'£—=—-=> :,=15_4m ,5; 120 {E {1}” )jc =[12_1)(1.35):> is =1.865mA p 3.5 3.6 “a I} h... ‘h‘ E 3 II 5. ‘6’: it *3 ﬁ U “In. II to :4 H1 3 :‘C = [ﬂ}21_7) = ic. = 27.4 In»! ' 27.7 is: ’5 =——=~ i5=0_304mA up 91 —— 3.7 DBVECB 12 is = Sle'ﬂp’f :3 05:10-3 = Sleuﬁsofunza So that [,1 = 6.94:10'” A Device 2: 12.2:10" = [522“sz Or 1‘,1 = Lame-'3 A I 1.69:10'“ . Ratio afareas = i: —--—::> Rang = 24.4 15. 6.94x10'” 3.3 r V (a) r,=—-"—=2—S—q:> rﬂ =250kﬂ Is I F 250 r=—"=—= =2‘SOMQ (b) ' 10 0.1 r“ 3.9 Bcha 50 V = = — 3 “9° an? m BVC = 12.9 V Electrgnic Circuit Analysis and Desigg, 2“" edition Solutions Manual 3.10 3.“ 3.12 BV BVcso= £0 -251 a -ﬂ- 56... W»ﬁ_ 56 _3.93 g=eo.s Bv BVch= \$38” 3v¢m=wvcmﬂ=mw€ﬁ I BVQE§=154V (a) 15:121‘00'7 => IE=1.13mA 75 I = — 1.13 =1.12mA c [76 3 I»:=5 = 24 '- (1.13)[10) —(1.12}Rc = a so that 1'?c = 5.98 kg 1 I =—=0.0132mA (b) , ,6 V, = —I,R, = —(o.0132)(so) => V, = 4.653 V Ic = [E 1) = 0.937 mA 76 5—2 =3“: 2 RC 0.937 L 1041‘“ c_ ;3=w:15=111w‘ 10 +[76)[10} I: = 0.905 mA V; = 0.7 +{o.0121)(10) — 2 V; = -l.lﬂ V V;- : IcRc — B = [O.906){3) — B = V: = -5.23 V Vic = V; — Vc = —1.15 -— (—5.25) n» Vgg = 4.1V :1. 5 =(1+ 3)!3(10} + 13(20) + 0.7 + (1 + BNBR} 5 = {BUN} +20 + 152] + 07 =- 13 = 4‘61pA' Ic- = 313 = US\$1.61; :- Ic = 0.346 mA V; = s — (1 + mime: = .5 — [TE)(0.00461}(10) = V; 21.50 3.13 (a) Figure P3.12(c) 3=(76}1,(10)+o.7+1,(I0)-—2 I = 10—0.? -' 10+(76)(10) Then ' 1C=(75)1,=a 1C=0306mA and 15:0.9I8mA = 0.0 l 208 mA Vac = 8' [5R5 " {CRC "('8) vac = 16—(0.918)[10)—(O.906){RC) vac = 6.32-—(0.906)RC RC: 31:91:593 = 2.355 RC glism Then 3.9? _<_ VEC S 4.24 V (b) Figure P3.12(d) 5=(l+ﬁ)I5RC+I,(20)+OJ +(l+ﬁ)15(2) 5 = (76)],RC + 1,00} + 0.7 + (76)IB(‘2} Now RC =10k£1i5%= 955 RC 5:051:31. Then 0.00443 5 I, ~_: 0.00481mA And VC =5—(1+ﬂ)1,RC So that L46 5 VC 1:153 v 3.14 V -v 2.5-0.7 R =M= =9 R =120kﬂ " I, 0.015 ‘—._-_—. Im=(70}(15,wl)=>1.05m v —v _ Rc=ﬂ=ﬂm RC=238kﬂ [ca 1.05 —— 3.15 _ _ -Va __ -(-1} a, V5——15R3=>IB— RB — 500 Is=2ﬂp5n ’Ez—l-CI.T=—ITV EVE—(n3)_—1.T+3_ 7 15.. RE .. I” -02 08mA 1E _ _ 0.2703 _ , _ _ E _. (1 +13) _ 0.002 —139.4 = ,3 — 134.4 .. 5 _ a—1+‘B=:'g—ﬂ.9926 1c =Bfa =9- Ic =0.259 mA Va: =3- VE=3—(-1.T}=>VCE :43? V Chagth 3: Problem Solutions 3J6 b. In: i¥=575=ﬂjmls 4 = 0.7 + 1335 + Us + chﬂc — 5 In + 1c = h; 4 = 0.7 + {3(100) +{0.5}(B} — 5 I 0.5 I; =o.o43=:. —E= (HI!) = ---— =11‘63 [5' 0.043 13:10.63, a: 13.6 =>g=ll9140 I. Va=0=>CutDﬁ=9fE=0,V¢-=EV -.7 b. V5=IV,I£=l ° :5 Ig=0.3mﬁ. fr: 2 I2 = Vc =5—{0.3](10)=> V}: = 3V c. V3 = 2 V. Assume active-mod: 2-10“=I£=1.3InA=Ic Vc = G — (1.3)(10) = —7 V1. Transistor in Man 2 -- 0.7 Is: f5: = = I; = 1.3 IDA Vs =1.3 V. ch(sa.t] = U32 V Vc = V; + Vcsisu] = J..3 + 0.2 =¢V£=1.5v b. V53=1V 1—0.7 I:— \$0 \$611!. fc=ﬂ[5=(75)(5)%fc=ﬁ.45mA S-Vo_ E s 'I°+10 1—0.45=Vo(%+ :Vg=1.33V c. Tmsislor in “Malina Va "—1 “cabin = 0.2 V 3.13 (a)/3p=100 . 100 (1) IQ=OimA 1c {Tb—l o.|)=o.0990 m _ V0=5—(O.099)(5):> V0=4505V 100 u = I = -— 0.5 =0.495MA (11) IE, 05mA c [101 ) I Va = 5 —(o_495)(5) : Va = 2525 V (iii) IQ = 2 mA Transistor is in saturation V0 = —VBE(sa!) + VCE(5ar] = —018 + 02 z} Va = —0.6 V (b) m = 150 . ' 150 (s) 1‘? = 011m 1}: = [ET 0.1) = 0.09934 m4 V0 = 5—(0.09934)(5) :5 Va = 4503 V Va change = Marlow/3 = -0.044% 4.503 .. 150 (u) 19 = 0.5 mA I: = a 0.5) = 0.4957 mA VD = 5—(0.4967)(S) 2) V0 = 2.517 V % change = Marlow. = —0 32% 2525 (iii) IQ = 2 mA Transistor in saturation Va = —0.6 V No change 3.19 I£=VEIOJ a 50 5—1/6 :2 = — V — = k ([+3)IE (51)“? 0'7) m IMVc=VE so . _6-V5 (51)”5 0'7)“ to 9.300}, - 0.7} = e — v5 10.30% = 5 +{n_7){9.501=. Va =1.19 v __ 1.19 — 0.7 13‘— —"'T""'-=>I5=0.49mﬂ. 3.20 S — 0.5 Vm=OjV=> VO=D.5V, 1c: =0.90mA 101 I = —— 0.90 = 9 [mo ):> IQ 0.909mA Electronic Circuit Analysig and Design, 2'“ edition Solutions Manual 3.21 10—V5_10-2 _ IE_ 10 _ m =>IE_G.80 mA '3=Vg—0.T=2—O.T=1.3V V3 13 f3=R—s=ﬁ=>lg=0.026mf\ [c 0.774 = .....= __ = "7 ﬂ 3 0.026 :5 29' 0 _ 29.7? -- .£I'—[_+a 307T=>a—096m V5: 2 V5 - vc = V5 - {IGRC — = 2 — [(0.774)[10} — 10] V5; = 4.26 V Load line developed assuming the Vﬂ voltage can change and the R, resistor is removed. I 11114 ._ -, 1;; 0.77% ‘9 39° " ﬂ; LL LG Vé‘: 2.6 3.22- so = — 1 = 0.90 mA 1,: [51} 1 Va = IcRc - 9 = (o.93)(4.7) — 9 or c = 4.39 V I, = lf=00195 mA Vit = 1.12, + V3001) = [0.0196)(50) + 0.7 or , = 1.68 V 3.23 so 05 = —— =0.49mA.1 =—=o.0093 A [c [51 05} , 51 :11 Va. = 1,11, +VE,(0n) = (0.0098)(50) +0.? or V‘. = 1.19 V Vc = c126 - 9 = (0.49)(4.7) — 9 = 4.701? Then V” = V, - V: =1.19—(—s.7) = 7.39 V P,2 = [CV55 +131!” = [O.49)(7.89) + (0.0093)(0.7) or Pa = 337 mW Power Supplied = P, = 19(9 4,) = (0.5)(9 — 1.19) Or P; = 3.91mi? 3.24 For IQ =0, then PD =0 For 1.2 = 05 mA , [C = [3}- 05) = 0.49 mA 51 I, = = 0.0093 mA, V, = 0.490V, V5 = 1.19 V VG = (0.49)(4.7) — 9 = 45.70 V 2:. Vﬁt = 7.39 V P 5 1?ch = (0.49)(7.39) :> P = 3.37 mW For IQ = 1.0 mA, Using the same caicuiations as above, we ﬁnd P = 5.95 2213’ For 12:15:71.4, P2626mW For19=2mA, P=4.80mW For 12:2.SmA, P=157mW For IQ = 3 111.4, Transistor is in saturation. 0.7 +I,(50) = 0.2 + 1601.7) — 9 IE =IQ=15+IC=II=3—IC Then, 0.7 +(3 — 1C)(50) = 0.2 + {C(17) — 9 Which yields 16 = 2.915 171/! and I, = 0.034 01.4 P = [EVE + 1’ch = (0.034)(0.7) +(2.916)(0.2) or P = 0.642 mW 3.25 IE1=I32=§=>IEI=IEQ=Dij I51 = Ic? z 0.5 mA Vc1 = Va: = 5 - [0.5){4) # VG: = Vgg = 3 V 3.26 a. [so = ——Hc -VB£(°n RB I59 = --—- = -a = 0.0333 01A 24—0.? Ra - 00333 = R3 — 599 kn Vcc -Vc£q _ 24—12 RC 1:» RC — 2 m-Rczﬁkﬂ log = _ Vac — V35(on) __ 24 - 0.7 1:. I39 —- RB — 699 = 0.0333 ILA (Unchanged) 1m 2 1015.: = [100)(00133) =? I'm; = 3.33 mA _. Vceq = Var: — Icoﬂc = 24 - (3-3)“) = Vega = 4.02 v M. gghaptcr 3: Problem Solutions (:3) was = Vcc - ICRC z 24 — [C(e) 3.27 IE = VEE —V53(on} = 9 —-D T as 4 => 1: = 2.075 M Ic = 1115 = [0.9920)(2.075) =19 IQ = 2.05 mA Vac +IcRc = Vcc '3: = 9 - [ZIUSNZJJ =5 VEE = 4.47 V 3.23 _ — Va; 12 — 5 Ice — T = = 2.73 mA [cg 2.73 [sex T: 3-0—3 Iﬁq =U.091mA 0.7— (-12) I = = R2 mu 0.127 mA In; = Im + {sq = 0.127 + 0.091: 0.315 1.11.“. V. = In: R. + 0.? =(u.21s)(15)+ 0.7 = V; = 3.97 V 3.2.9 For Veg = 1.5 5 —‘l.5 Icq = I = 0.5 mA 0.5 [sq _. Eg— = 0.02 IRA 0." — — 133 = —‘Tgsl = 0.057 mA 1m = {R1 + 19:: = 0-057 + 0.02 = 0.077 mA V. = In; R. + Vas(on) = (0.01?){15) + 0.1 =1.35V FWch=1.0 1cq=a:1=-imA 4 I =-——=0.6 an 25 1 nut [33 = 0.057 M Im = In, +I5q = 0.057 + 0.16 = 0.217 mA V. = {0.2171(15) + 0.7 =- 3.96 v So 1.56 < V1 < 3.96 V 35:. q. ‘1' (544%.; or: Q'fA‘. VJLUEJ; a; -— o 1 «ar- y 3.30 R“, = R1IIR; = 33||m = 7.67 m R: . ID V” ‘ (R. + 3.)“ ‘(1o+33){15} = 4.19 \r' VT}: — V35(Dn) _ 4.19 - 0.7 Rm +[1+ m3: _ 7.5? +(51)(1) fag = 0.0595 1111’; Leg = 160 = ISIBQ ‘4' Ice = 2.97 mA [sq = 3.03 mA Vcsq = 'cc — Ichc - 1'squ = 1a . [2.97)(2.2) — (3.os)(1; :Zsﬂiﬂﬂ 3.31 ImzllmA, ch=9v, Rm=som Also I,=-§-g~=0.015mA V”, = {mm +V,E[on) +[l +ﬁ]IaQRE V”! = [Eff—R1}Vcc) = Rm 'Vcc = "i"{so)(13) Then éﬁoxm) = (omsxso) + 0.7 +(31)(o.015)(1) - 338R. - or R, 4538“). Then m-SO: R, =ss.7kn 15:, {\$3 1.2) = 12151111! 18=Ich +Vm+meE 18 =(12)Rc +9+(1215)(1}=> RC = 6.49 m Elcctronic Circuit Analygis and Design, 221"d edition Solutions Manual 3.32 v” = (&){24} = 9.5 v 39.9+60.9 R,,,=R.|;R,=20ﬁ15=057m 95 D_ __. —' ‘ " = 0.00553 A V = R? V )=[ 15 )(lolmtzw I“ 24.1+{126}(10.5} m m R. +R: cc 15+20 [cg = 0.317 mA, 13.; = 0.823 mA I = _. r v =15.w Vac ___ I.ng +Vﬂ(on] + £9 +Vm VCEQ 24 . (0323)(10 a} 3 can 50 0.817 5 [cg 51.02 mA and 10: rm(1)+0.7+fm[al'ﬁ]+4,29 14.2 g ch g 10.1 v. The“ 3.34 -O.7— . . R =R R =258=6.06 in I =10 429:5 l:> I =4‘62m‘4 3 TH 1|] 1 I] “9 +331 1.035 ia____ V _ a? )V _ s \[m 101 m“ R1+R2 m" s+25 I . . ﬁll-Rm +V;W=[1§‘i 357)+4.29 =5.52\ 1+}? 101 I _ a”, “magma; = 5.52 — 0.7 0r 3‘3 ‘ Rm +(1+ am; 5.05 +{76)(1] ml: 13¢ = 0.0524 mA. Icq = 4.50 mA 3.33 159 = 4.74 a. Rn; .= mum = 531142 = 24.35 km Vcsa = Vcc - IcaRc - Isa Re : '2 — . 3 - 4.74 '1 V _ R: \V y 42 24 i .4 man: i JH 7"" 3:4:sz CC“ 42+sa ( ) ___"CEQ=°-12V = 10.1 v 5.52 — 0.7 b_ r = —-———— I = [1.0326 mA I _ Fm — Vsﬂon) __ 10.1- 0.? 5" 5.06 + (151x11 : i"— B" “ Rn: + (1 + 5mg "‘ 24.35 + (126)(10) I“, = 4.39 talk by; = 0.00732 mA [sq = L92 1:0 = M80 ={125M0-00732J chq = 24 — (4.39)(a) — (4.92m) => [co = 0.915 [11A chq = 4.41.“. 15.; = 0.922 a chq = 24 — (0.9223(10) 3 35 =>Vc50=14.sv (a) [WEIRQ=0‘4MA a. Let Rc=i=>12c=15m; RE=—3—:>RE=7.5kQ R2 = 4:2 ta + 5% = 44.1 m 0.4 0.4 9 R:=58kﬂ-5%=55.l kn + 5—.—=1125k{1 R‘ R’ (0.2)(0.4} R5 =10kn—5%=9.5kn Rru = min: = 24.5 m pm =[RiiiRjyoJ = [593,” +V,E(on)+(1+ﬁ)I,QRE «.1 = —.__ = , 112.5— Vm (“'1+55.1)(24) 107v Rm Rd?1 _( Rama, 10.7—0: "R,+R,_ 1l25 LEG = =0.00319 I =0i=oﬂo4m log = 1.02 mA. I“ =1.03 mA ’9 100 ) . [125—111 R, VCEQ -"'- 24 — (1.03)(9.5] => chq = 14.2 V 9 ___ (_______,_ +0]! R: 112.5 (0.004) 112.5 R L“ m y n 4100(000405) = {2 -— = . a 5 o 39 g k We obtain R5 = w m + 5% = 10's m From this quadratic, wcﬂnd Rrazﬂ1llﬂz=24.lkﬂ R1=48kQ =2! R1=545m Chapter 3: Problem Solutions 0)) Stande resistor values: -Sct RIr =Rc =15H1 and R, =62kS1 R, =47m Now R". = RallR‘J = 62||47 = 2mm _ R2 _ 47 _ Vm {RI +R1)(Vm]-[4?+62)(9)_ 3.331! Vm = [IQRTH +V2£(O"}+{1+ﬂ)laQRr So I — 338‘“ —000406 A ’9 26.7+(101)(7.s)" m Then 49:03:06)“ pm = V“ =(0.406)(7.5)= 3.05 V 3.36 1- RTE = Kill-R: =12"? = 1.71 = HIE . R2 2 v = — = _ ’ TH (R1+R2)(10) 5 (12+2)(10) a = -—3.ST V = V?” b. V —- _. _ 13‘; = TH Vasfon) ( 5) Rn; + [l +13)R£ _ —3.57 .— 0.7 + s 0.73 - "" =- Iaq = 0.0140 111A Ice = 1.40 mA, Isa -'= 1.41 mA. chq =10 -- Ichc- - IEQRE =10 — (1.401(5) - (1.41)(0.5) ch = 2.30 v d.For R: =2kn+5%=2.l m R: =12 kﬂ-5%=11.4m R5 = 0-5 m — 5% = 0.475 m Hrs = Rallﬁa = 1.7? tr! 2.1 V73 -- (2.1 +1I‘)(10) -5 -- -—3.44 V —-3. —0.7 5 . I“ __ . ____........_4" + — 9—55 =0.0173 mA ' 1.77+{101)(0.§753 " 49.7 1.3;; = 1.73 mA. {sq = 1.T5 mA F01ch =5kﬂ+i%=5.25kﬂ Vcso =10 — (1.73)(5.25) — (1.1301475) =0 V539 = 0.0863 V (Saturation) F0! R2 =2kn-5%=L5m Rl = 12 kﬂ+5% =12.6 kn R5 = 0.5 kn + 5% = 0.525 kn an = R; "R: =1.55 kn 1.9 V = —_ ﬂ _ a _ _ V T” (12.s+1.9){1 ) 5 3 69 —-3.69—0.7+5 0.61 I = — = —-- = , 3" 1.65 + {101)(0.525) 54.7 0 0112 I“ [co = 1.12 mA. Isa = 1.13- In}. For R;- =5 kﬂ-S‘FE =i.75 kn Vcaq =10 - (112101.75) - (1.13)(0.525] =-.~ Vcaq = 4.09 v \$01.12 5 leg 5 1.71 mA and 0.0863 (V < . V. _ cgq _ 409 Saturation 3.37 RTH = R1HR2 = 9H1: 0.90 kn Vm =(R1‘ERQ)(-12)=(llﬁ)(~m = #1.? V IBQRE + Vgaﬁm) + [Haﬁz-H 4- V7}; = 0 Mn.) _ 1-2 -' 0-7 Rm +(1 +0135 ‘ 0.90 + {76)[0.X) [30 = 0.9533. Icq = 4.41mA Lac = 4.47 mA fag: Cantuaflmdlinc: Vgcq 26V 1293-: + Vscq + h:ch- -12 = U (4.17)(0.l) +6 + (unm- =12 av R; = 1.26 kn Electronic Circuit Analysis and Design, 2"" edition Solutions Manual 3.33 (a) R”, = (0.1)(1 +133; = (0.:)(101)(05) = 5.05 kﬂ 1 VT" = K. Rm .0“ = 1,077," +V,£(an)+ (l +ﬁ)7,QRE l I . ,r —.£G_=ﬁ=o,oosm ‘9' 13 100 Then \$505710) = (0.003)(5.05)+ 0.7 +(l01)(0.008)(05) OI.’ .R RI =44'Em‘ £41; 44.! + R2 New {£9 = [191)(03) = 0.303 m 100 vac = Ices: Wm + 1593. 10 = (0.8)RC + 5+ (0303)(05) RC = 5.75 k9. = 5.05: 132 = 5.70 7.9 (b) For 75\$ﬁS150 _ R, _ 5.7 _ vm {RI “a: )(VCC) _ (—-—5‘7+M'1](10}- u45v VT" = “QR”, +V,£(0n)+(l +13)I3QRE _ l.l45—D.7 ‘9 7 5.05+(76)(05) Then [CO = (75)(0.0703) = 0.775 m For 13:75. I =0.0103mA 1.145-0.7 For = . I = = ﬂ ’9 5.05+(151)(05) 00552 “A Then [CE = 0.829 m AI .. 95 Change = m = 0529 0375 x100% =- Icq 0.80 96 Change = 6.75% (a) For R5. =15] Rm =(0.1)(10i)(1)= 10.1730 Vm =“Rl—' Rm 'Voc =%‘(10.le0) =- l (0.003)(10.t}+0.7 +(10:)(0.003)(1) which yields RI = 63.6 kg 63.617, 63.6 + R2 12 NOWV = R2 V = = ’" [almjcc] [12+63.6](10) '58” 1587—03 F =75. I = = _ And =10.1==. R, = 12.0139 So IQ = 0.773 m l537 - 0.7 FD =15 II = “6 o '9 10.1+(151)(1) = 0.0055! mil 3.39 3.40 Then Ice = 0.826 M Mm _ 0326 — 0.773 [W ‘70 Change = 6.63% % Change = x500% =3 Vac a Immc + RE)+ch 10 = (0.8)[Rc +R£)+5= Rr + RE =6.25 7.9 Let RE =1kﬂ Then. for bias stable Rm =(0.l)(121)(l) = 12.1 kﬂ 0.3 I” = m = 0.0066? 77174 \$41ti10) = (0.00667)(i 2.1)+ 0.7 I +(121)(0.00667)(1] 76.2771 30 R, = 76.2 m and —— 76.2 + 772 212.1: R1 = [4.47.0 Then 1’ "Hm-Lo— ” ‘ 76.2+14.4 This is close to the design specification. =0.l 10:77.4 Ice = Lee = cho = Vcc - Icqlﬁc + RE) 3 = 12 — Icqm + 0.2) 1.5:; = 2.73 31A. 13¢ = 0.0213 mA Veg-q = a v VIN = IEQRTH + Vaalon) + (1 + ﬁNaQRg -6 . R: ) l? = 12 - 5, R = R R TH (Ri +32 ( ) TH I“ : Bias smblc=> RT}; = (0.1)“ + 3)]{5 = (0.1)[125)(D.2} = 2.52 kn my = (3%?)(RTHX12J-6 i(2.52)(12) — 3 =(0.0213)[2.52} + 0.7 R: + (123)(0.0213}{0.2) — 3 343024) = 0.7549 + 0.5494 R1 23.2122 — ° '1 —- = 2." M- 213+}?! 32 R2 = 2.33 kn _. _ _ ghaptcr 3: Problem Solutions 3.41 80 I. Ice = 1 mA. 15:; = )(1) = 1.01 mA VCEQ =12 - "- =D VEEQ = 9.30 V 15,; = 31—6 = 0.0125 mA R” = +[o.1}{1 +5112; =(o.1)[31')(o.2)=1.62 m R: _ l a _ P'ng (RI+R:){12)—6—R—I(RTH)(i-) E 1 _ R—1(19.44}—- 5 VT” = {EQRTH + VBE(OI§) + (1 + 15)IEO RE "" 6 RLuwm) - s = (0.0125)(1.62) + 0.1 1 +(31)(o.0125)(o.2) — a ill—(19.44) =o.923 1 _ 2!.132 R! - 21.1“), -- 1.52 R; = 1.75 kn b. R; = 22.2 1:9 or R1 = 20.0 H"! R: =1.“ 1:0 or R; = 1.66 kn R5; = 0.21 1:5"! or R; = 0.19 kn RC = 2.1 kn or RC =1.9 kn Rﬁmu}. Rdmin). R5(m‘m) Rn.- = (1.34}:|(2n.0) = 1.685 m 1.84 V _ _.._._ _ — _. TH (LB‘H_ 20_o)(12} 6 4.99 V 4.99 —o.7 — (—6} _ 0.31 1'.635+(81)(0.19) ‘ 17.05 Icq = 1.45 mA For max. 12.: a Veg = 12 -—- (l.45)(2.1) - (1.47)(D.19) VCE = 3.63 v Isa = = 0.0182 nm R2(nﬁnJv R1(ma~x)l R“,- = (1.sa)||(22.23 = 1.547 m 1.66 v = _ _ =_- _ T" (1.66+22.2)(12} 5 a-lﬁa'v’ -5.155— 0.7+6 0.135 r = ———_.__ = __ 3" 1.547 +(31)(o.21) 15.55 = 0.00727 11A For min. RC ==» 159 = 0.53: mA, I; = 0-539 Vcsq = 12 - (o.552)(1.9) — (o.sas)(o.21) chq = 1037 V So 0.532 _<_ Ic 5 1.45 m 3.65 S Vcsq 5 10.77 V 3 .42 3.43 Vm2 Eva —IW(RC +RE) 5: I2—3[Rc+RE)=> RC+R£ 22.3316). but R5 =0.33m and RC =2m Nominal value of A3: 100 Rm ={0.1)(1+ @125 = (o.1)(101)(0.33) = 3.33 L9 [B9 =%0=0.03m.4 l l .. pm =E. Rm .92)- 5: ET(;.33)(12)— 6 Then Vm = 1,91?“ +Vn[on)+(l + ﬁNBQRE — 6 Ri(3.33)(12)— 6 = (0.03)(3.3 3) + 0.7 + (101)(o.03](o.3 3) — 6 which yields R. = 22.2 kg and R1 = 3.92 kﬂ Now R 3.92 V _—_ 1 12 -6: ——.__ 12_ ”' [3%ng ) [1924422) 3' 6 Or VT" = 4.20 V For ﬂ = 75 , Vm z IsaRrH +Vx£{on)+ (1 +16)IBQR5 ‘ 6 I = vm +6—0.7 = wane—0.7 ’9 Rm + (1 + [3)125 3.33 +{76)(033) = 0.0387 m =5 15 = 2.90 m For 15‘ = 150 . I = 4.24-6—03 "3 333+(151)(o.33) Then IC =3.10mA Spaciﬁcations arc met. = 0.020? M Rm = RlﬂR, = 3&1: = 2.4 m R1 _ 12 _ Vm=[R1+R1)Vcc_[12+3](20)—16V (a) For {3:75 20 =(1+ muck. + vﬂogn) + I” R," + v”, 20— 0.7 — 15 = I,Q[(?6)(2) + 2.4] 50 1,9 =0.0214 mA, {CE = [.60 mA. rm =15: mA Vm =20-(l.6)(1)—(1.62)(2) 01' vm = 15.16 v (b)Forﬁ=100,wcﬁnd I,Q=0.OI6|nm. Icazmlm. Vs =IS.I3V Electronic Circuit Analysis and DesignI 2"" edition Solutions Manual 3.44 Icy;- =4.B mA - 15;; = 4.84 mA Vase = Vac - Icoﬂc - 15133.5 5 =13 — (4.311(2) — (4.34m; => RE = 0.496 H1 RT" = (0.1“). + ,6)R£ =(0JHI21MOAQG) = 6.0 kn Vrg = IBQRTH + VBE(°“) + [1 + ENBORE‘ l I V = .._. R .V = _ _ TH R1 TH cc- Rlﬂi CHIS) 31—10mm =(o.04){s.0) + 0.70 + (121){o.04)(0.496) 1 if{(1011) = 3.34 32.31%; = . kﬂ. —--—- = . 33—33—- 320 + R: 6 0 R: = 7.37 kﬂ 3.45 For nominal .3 = T "J ng = ﬁ = 0.0235 mA - [so = 2.03 m."|. Vcsa = Vc'c: - Ichc -- 15133.5: 10 = 20 — (2)14) — {2.03335 9 RE = 0.935 K RTE = (0.1)“. + ﬂute = (0.1}(71}{D.955} = 6.99 K V73 = ngRrH + ng(on) + ngﬁg 1 .. _ R—I - Rn: - 'ch = 5205171; + V5503“) + 15035 Ril(5.99)(20) =(0.0236)(6.991 + 0.70 + (2.033(0935) 1 R71139.11} _. 2.90 \$8.21!: R = _2 ‘ _._.._ = _ -‘————“ K 43.2 + R, 6 99 R3 = 3.18 K Chuck: For .3 = 50 1.13 v _—, _.__ = m (“a +“Hg-Jan) 2.90 Vm - Vagina) _ 2.90 — 0.7 I = —-——————— _ _._.._..___ 9" Rn: + (1 +533; 5.99 + (511(0935) = 0.0334 mm Icq = 1.92 mA For )5 = 90 La - ﬂ — o 0223 A ° ‘ c.99+(91)(0.935) ‘ ' m 1:2 = 2.05 mA D. ..nu.al 3.45 [cg =1 mA at ng =1.02 mA V's-sq = Vcc - Ichc - IEQRE. 5 =15 — (1)15} — (1.02135 => RE = 4.90 kn Bias stable: - RT” = (0.1)(1 +3)R5 == (0.1)(61](4.9] = 29.9 kn 1 {HQ = 63 = 0.0167 mA 1 . VTH .._. ETH- - Vac = IBQRTH + I"BE{cm') +IEC|RE Tel-(29.91115) 40015702901 + 0.70 1 + (1.02)(4.901 Laws) = 5.197 R1 72.43: _ R] - 72.4 kn, 711+ R7 — 29.9 R: = 50.9 kn CheckrFo:5:-.45 50.9 v = ——--- ’ =. v 7" (50-9+72.4)”°) “9 WI'H -V3£(on) 6.19 —0.T 13° = m = m = 0.0215 0131 Icq = 0.953 319.. 91-? = 3.27. Chad: For .3 = 75 Ian = mm— = 0.0136 m2. 29.9 + (75)(4.90) Ice =10: mA. —° -.-.- 2.0% It: Design crilcrion is satisﬁed. 3.47 (a) Vcc 5 1.31025 + R£)+vm 3 = (0.1)(512'E +RE)+1_4 :9 R5 = 2.57 11': RC = 13.319 , 1,0 =%=0333m R,” = (0.1)(1+ ma. =(0.l)(121)(2.67) = 32.319 1 1 Vm = F'RTH "vac =F(32'3)(3) I I = “QR”, +V,£(on)+(1+ ﬂ)I,QRE = (0.000833)(31.3)+ 0.7 + (121X0.00083 3X16?) which gives Rl = 973 El, and R2 =4S.4 m ___—_..__ Chagter 3: Problem Solutions 3.48 3 .49 3.50 3 3 (b) 1',[ E R1 HQ: zm=>20ﬁm rm=1mm P = (im +1.11% =(l00+?.0.6)(3) 01' P=362 ,uW 1525—;‘E-§=1.67m R5 - 3 RT}! = R1IIR: = (0.1}(1 + )3}RE = (0.1](101)(3) = 30.3 k0 R: 1 v73-(R1+R2)(4)—2-E'RTH-(4)—2 Isa 1+5 I50 = = 0.0165 nut 5 = ngR3 + V2.9(on) + [BETH + V”: 5 = (1.5T}{3] + 0.7 + [0.0165)(30.3) 1 + R—‘(30.3}[1) - 2 0.30 = \$430.3)“; => 5:, :15: m 1 152R; _— = {13 = _ 152+R: 3 = R3 3781:!) 5- Rm = Rle-R: =10[|20 =5 Rm = 6.67 m . _ R: 20 1'” " (It. +Rg)(1°) ‘5 = (20 +1::h)(ml ‘5 =:v VIE = 1.57 V b. 10=II+BJIBQIR£+VE£(OII)+IEQRrH-FVTH I3 = 10—03—157 _ 7.53 ° 5.s7+[51)(2) _ 123.? => Iaq = 0.0593 LuA Ice = 3.65 EDA, Isa = 3.52 mA V5 = to - 15°35 =10 — {3.52)(2) V; = 2.76 V V: = ICQRC — 10 = (3.56)(2.2) — 10 'c = -2.17 v V‘ -V‘ :-: [camc -t-RE)+V5m 20: (OSXRc + R£)+8=:(RC+R£)= 24 in Let RE=10kﬂ than Rc=l4kﬂ be! ,8 = 60 from provious problcm. R1." = (0.1)(l “Sm. = (0.1)(61}(10) 351 0r Rm=61kﬂ 1,0 =% ='O.DOS33mA R 1 V = 2 l0 -5=-—-R -10-5 m [ R1 + R2} J R‘ m Now IO = (1+ﬁ)!80RE +V55(0")+ Irmg‘zem +VTH 10 = (61)[0.00833)(10) + 0.7 + (0.00333)(61) l +— 61 10 —5 RI( X ) Then R,=70.0k.Q and 31:47419 - 10 _‘ l0 Rl + R2 70 + 474 So the 40 1% current limit is met. In Ill :alSAM L a”. = mug, = 351120 => 13:12.7 kﬂ R: _ _ 20 7_- VrH=(RI+R2)(‘)—5-(20+35)(} 3 => VT}; = —2.45 V 13. V71; - Wash“) -[-10) RT}; + (1 + MR: —2.45 - 0.7 + 10 " 12.7 + (76)(D.5] =- Im; = 0.135 \$139.. Lea = 10.3- In). I59: ICE 2 10.1 m. chq = 20 — J’ch; — {was = in — (10.1)[0.3) — {10.3)(u.5) vcm = 5.77 v C. R: =20+5%=21kﬂ R1 = 35 - 5% = 33.25 kn R3 = 0.5 — 5% = OATS k0 R1”: = RﬁlR: = 2133.25 =12.9 kn 3: V-rH— (R1+R:)[7} 5 2] 2 (21+ 33.25)ml " 5 " ‘2'" V —2.29 - 0.7 - (—10) W Icq =10.T mA. [sq =10.9 111A 15¢: =D.143mA For R:- = 0.3 + 5% = 0.84 kn cho = 20 - [1o.7)(u.34} - {10.9}(o.475) : chq = 5.5-3 V Electronic Circuit Analysis and Design. '2"d edition Solutions Manual For Rc = 0.8 - 5% = 0.75 kn Vase = 20 - (10.7}(035) — {10.9}(o.475) =3- Vcsq = 6.69 V R1=30-5%=l9kn R: = 35 + 55’: = 36.75 kn R9 = 0.5 + 5% = 0.525 kn R”; = {7.1”}!2 =19|i36.75 =12.s m 19 VI. _— _._..._.._ 7 _ - _ _2‘ v H (19+35.75)() 3' 61 —2.t‘:l —n.7 - (an) 12.5 + (757(05sz [ca 2 m5. [£0 = mA {sq m = 0.128 mA For R: = 0.84 kn VCEQ = 20 - (9.53)(0.84) - (9.703(0525} 3 chq = 6.36 V For H: = 0.76 kn Vc'gq = 20 - (9.55)(o.76) — (9.70)[0.525) : Vcsq = 7.63 V 50 9.53 g Icn S 10.7 mm and 5.83 5 chq 5 7.63 V 3.52 L an, = sun musoo mum m = 256 mum m =7? REE = 5%.? 160 .5“er 3-VTH_VTg—!-S) 500 500 ' 7o 5 3 5 1 1 1 500 + 500 - 7n -"”(§7+aﬁ+ a) -- 0.0554 = VTH(O70153} VIE = —3.03 V b. [so = VTH -V55(an) -- (-5) R73 = (1 + #3135 _ 4.03 — 0.7 + s ' 54.7 +{101)[5} Isa = 0.00227 mA Ice = 0.227 mA. 15¢ = 0.229 Page = 20 - (0.227)(50) - (0.2mm cho = 7.51 v 3.53 3.54 Rn; = JDIIGDQIQU =- Rzﬂ =10 kn 5-VTH 5-V7'H =VTH 30 60 2D 5+5 _V 1+1+1 30 so ‘ T” 30 so 20 ’1 For 3 = 100 V7}: - Vasfon] — {—5) RTE + {1 + {31195 _ 2.5 0.? 4-5 ‘ 10 + (101M031 [sq = FHA. [Cg = 32.5 mA. [.59 = 32.7 m}. Vcsq =15 — (22.5]10.5) — 22.7mm) In sanitation ng= VCEQ = ~0.79! =3- VCEQ = 0.2 V Vs = VT}! - IBQRTH - VEE{°'1) = 2.5 -- (0.225H10) - 0.7 ’5 = —U.45 V :- Vc = -U.I15+ 0.2 = —0.'L'5 V 10 — (—0.25; I” = 0.5 =>' Icq = 20.5 mm Rm = R: HR: = 100nm = 23.5 m R2 40 v _ — = _ .. _ TH (R;+R:)(m] (4ﬁ+1no)(m) 286V Vm - V32{on) 2.35 - 0.7 Rm +(1 Hanan = 23.5 + (1mm IE. = 0.0144 mA I37. = IQ, 21.73 wt, IE; =1.75 mA 10 — v —3—52 = In + In: I _ V5; — VEEODDJ — (—10) E1 — quh—‘F' lﬂ—Vag _I V32—0.7+10 3 ' “ (121)(5) m 9.3 1 1 — — .7 - — = v - 3 1 3 605 3’ (a + (121m)) 1.59 = Vm(o.3as) =3» V3, = 4.75 v 1.75 — 0.7 — (“10] 152= 5 =Iﬂ=2.31mA IE; = 0.0232 mA 1;; = 2.79'mA Vase: = 4.75 - {1.75)(1) =3 V5301 = 3.0 v chq2 = 10 — (-4.75 - 0.7) = Vega: = 5.95 V —-————-—————-——-_.___._. Chapter 3: Problem Solutions 3-55 3.5? RT}; = R1 "R: = SUIIIDD = 33.3 kn .. _._ R: lem=-o.T—.07=—14 “1+3” _].4_[_5 ‘ m 100 ) _ _ 7v I52="_1_:I§1=3.5 mA _(100+50 (In) 5—1'5 gm = 0.04.“ mA 5 = [51321 + VEB(°n) '4’ I51 RTE + VTH In 2 3.50 mA 10] = — . = 0.503 IE. (my) a) m 151: 1}“ + 151 = 13! = [ILA 151 = 0.259 mA 5 = (9.505333! + 0.7 + [0.008}{33.3) + 1.67 IE; = 0.00320 mA R5, = 2 93 IL“ [:1 = 0.256 mﬁ. vi, = s - (0.003)(2.53} = 2.63 v 3.56 V51 = V31 - Vgcq: = 2.63 - 3.5 =2 -0.37 V Current through 1" source 2 15: + £2: and V“ = “0‘87 _ 0'70 = 4‘57 v _ "7... — I51 = 12:2 = (1+ BUB: = (51)(B.26} pA 13; = iris-i]— =0.803 :9 RE; = 4.25 kﬂ So total cum-.01: 2(51)(B.26) 4023. = 843 uA P' = I: IV'| = (0.3mm => ' = 4.22 mw Vcam = 4 => ch = -1-57 + 4 = 2-43 V (me V" source) RC: = 5 31'“ =:~ RE, = 3.21 m From Example 3.15.14, = 0413 um IRm = Ic, - 15.: = 0.0 — 0.003 = 0.792 111A _ 50 _ —0.07—(—5} _ 50 Ice —- (51)(0.413) = 0.405 mA Rm - —-——-D_m =- M P" = I - W = (0.405}(5) => 2+ = 2 03 mW (From V+ sauna) ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 19

chap003 - .EICCtronic Circuit Analysis and Design...

This preview shows document pages 1 - 19. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online